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1) Why is the centre of the circle:

(x + a)² + (y + b)² = c²

(-a,-b) ?


2) What is a good way of remembering that the centre of the circle:

(x + a)² + (y + b)² = c²

is (-a,-b) and not (a,b) ?

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  • $\begingroup$ The center of $(x-(-a))^2+(y-(-b))^2=c^2$ is $(-a,-b)$, just as the center of $(x-a)^2+(y-b)^2=c^2$ is $(a,b)$ $\endgroup$ – J. W. Tanner Apr 16 at 20:45
  • $\begingroup$ Right. Why is that true? $\endgroup$ – Tom Apr 16 at 20:45
  • $\begingroup$ distance from $(x,y)$ to $(a,b)$ is $c$ $\endgroup$ – J. W. Tanner Apr 16 at 20:47
  • $\begingroup$ It might be useful to remember that in just one dimension, the distance between two points on the number line is (the absolute value of) their difference, not their sum. For instance, if you have a point at $a = -2$, and another point at $b = 4$, then the distance between them is $|a - b| = |4 - (-2)| = 6$, not $|4 + (-2)| = 2$. In fact, we could express the distance $c$ between them using the equation $c^2 = (a-b)^2$, and it would be perfectly valid. We only normally don't do that because in one dimension, it isn't necessary. But it would make the generalization to $n$ dimensions clearer. $\endgroup$ – Brian Tung Apr 16 at 20:55
  • $\begingroup$ that makes sense. thanks! $\endgroup$ – Tom Apr 16 at 20:56
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The equation relating points lying on the circle centre $(a,b)$ and radius $c$ is given by $$\sqrt{(x-a)^2+(y-b)^2}=c$$ where the left hand side gives the distance of the point $(x,y)$ from the centre and the right hand side gives the radius. For the distance from point $(x,y)$ the centre to be equal to $0$ we need $(x,y)=(a,b)$ hence the coordinates of the centre must be $(a,b)$. The equation above can be rewritten as $$(x-a)^2+(y-b)^2=c^2$$ $$(x+(-a))^2+(y+(-b))^2=c^2$$ which gives the same form you mention.

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It might be useful to remember that in just one dimension, the distance between two points on the number line is (the absolute value of) their difference, not their sum. For instance, if you have a point at $a = -2$, and another point at $b = 4$, then the distance between them is $|b - a| = |4 - (-2)| = 6$, not $|4 + (-2)| = 2$.

enter image description here

In fact, we could express the distance $c$ between them using the equation $c = \sqrt{(a-b)^2}$, and it would be perfectly valid. We only normally don't do that because in one dimension, it isn't necessary. But it would make the generalization to $n$ dimensions clearer.

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I would instead say that the centre of circle $$(x-a)^2+(y-b)^2=c^2$$ is $(a,b)$.

Notice how this substitution requires $c^2=0\implies c=0$.

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    $\begingroup$ I don't understand this. The radius of the circle is $(a, b)$? Do you mean that if one considers the degenerate "circle" that consists of just that point $(a, b)$, the radius is required to be $0$? $\endgroup$ – Brian Tung Apr 16 at 20:56
  • $\begingroup$ You shouldn't, because I'm an idiot who can't proofread. Corrected $\endgroup$ – Rhys Hughes Apr 16 at 20:58
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    $\begingroup$ And yes, that is the conclusion I am drawing ultimately $\endgroup$ – Rhys Hughes Apr 16 at 20:59
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The euclidean distance from the point $M(x,y)$ to the point $C(a,b)$ is the length of the vector $$\overrightarrow{CM}=\overrightarrow{OM}-\overrightarrow{OC}=x\,\vec i+y\,\vec j-(a\,\vec i+b\,\vec j)=(x-a)\,\vec i+(y-b)\,\vec j;$$ whence the formula for the length in an orthonormal basis.

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  • $\begingroup$ The brave anonymous downvoter was lurking around here again! $\endgroup$ – Bernard Apr 16 at 21:20

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